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['A',1]=>1 ['A',2]=>2 ['B',2]=>2 ['G',2]=>2 ['A',3]=>16 ['B',3]=>42 ['C',3]=>42 ['A',4]=>768 ['B',4]=>24024 ['C',4]=>24024 ['D',4]=>2316 ['F',4]=>2144892 ['A',5]=>292864 ['B',5]=>701149020 ['C',5]=>701149020 ['D',5]=>12985968 ['A',6]=>1100742656
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The number of reduced decompositions of the longest element of the Weyl group of the given Cartan type.
Equivalently, this is the number of chains in the weak order from the identity to the longest element.
In type $A_n$, this is
$$ \binom{n+1}{2}!/(1^n 3^{n-1} \dots (2n-1)^1). $$
In type $B_n$ and $C_n$ this is
$$ (n^2)!\prod_{k=1}^{n-1} k! / \prod_{k=n}^{2n-1} k!. $$
[1] Stanley, R. P. On the number of reduced decompositions of elements of Coxeter groups MathSciNet:0782057
[2] Kraƛkiewicz, W. Reduced decompositions in Weyl groups MathSciNet:1330543
[3] Winkelman, B. Number of reduced decompositions of the longest element of the Weyl group MathOverflow:370333
def statistic(c):
    if c.type() == "A":
        n = c.rank()
        return binomial(n+1, 2).factorial()/prod((2*i-1)^(n+1-i) for i in range(1,n+1))
    if c.type() in ["B", "C"]:
        n = c.rank()
        return ZZ(n^2).factorial()*prod(ZZ(k).factorial() for k in range(1, n))/ prod(ZZ(k).factorial() for k in range(n, 2*n))
    P = WeylGroup(c).weak_poset()
    p = P.chain_polynomial()
    return p.leading_coefficient()

# alternative, slow implementation
def statistic(c):
    return len(WeylGroup(c).long_element().reduced_words())
Aug 28, 2020 at 19:40 by Martin Rubey
Aug 28, 2020 at 22:18 by Martin Rubey